Question

Solve sec(2x) − 3 = 0 for the four smallest positive solutions


Thank you!

Answer 1

Answer:

$x_{1,0} = 0.196\pi$, $x_{2,0} = 0.804\pi$, $x_{1,1} = 2.196\pi$, $x_{2,1} = 2.804\pi$

Step-by-step explanation:

The trigonometric equations needs to be rearranged in term of one fundamental trigonometric function:

$\sec 2x - 3 = 0$

$\sec 2x = 3$

$\frac{1}{\cos 2x} = 3$

$\cos 2x = \frac{1}{3}$

$\cos^{2}x - \sin^{2}x = \frac{1}{3}$

$1 - 2\cdot \sin^{2}x = \frac{1}{3}$

$2\cdot \sin^{2} x = \frac{2}{3}$

$\sin^{2}x = \frac{1}{3}$

$\sin x = \frac{1}{\sqrt{3}}$

$\sin x = \frac{\sqrt{3}}{3}$

$x = \sin^{-1} \frac{\sqrt{3}}{3}$

The values of x are contained in the following two sets:

$x_{1} = 0.196\pi + 2\pi \cdot i,\forall i \in \mathbb{N}_{O}$

$x_{2} = 0.804\pi + 2\pi \cdot i,\forall i \in \mathbb{N}_{O}$

The first four positive solutions are:

$x_{1,0} = 0.196\pi$, $x_{2,0} = 0.804\pi$, $x_{1,1} = 2.196\pi$, $x_{2,1} = 2.804\pi$